189
Acta Math. Univ. Comenianae
Vol. LXI, 2(1992), pp. 189–192
ON THE VOLUME OF THE
DOUBLE STOCHASTIC MATRICES
M. SCHMUCKENSCHLÄGER
1. Introduction and Notation
Let
Q
n
be the group of permutation matrices in Rn . Then
Dn : = co Π : Π ∈
!
Y
n
2
is a convex set in Rn . It is well known that Dn is the set of all double stochastic
matrices, i.e.




X
X
Dn = T = (tij )ni,j=1 :
tij =
tij = 1 ∀ i, j ∈ {1, . . . , n}, tij ∈ [0, 1]


i=1
j=1
The volume of Dn is somehow related to a Kahane type inequality (cf. [S]) for the
group of permutations, more precisely: let (xj,k ) be a double sequence in some
Banach space X, if
1
√
Volk (Dn ) k
≥c n
2
Volk (Bk )
2
where k is the dimension of Dn ⊆ Rn , then the L1 -norm and the norm associated
P
with ψ1 (t): = et −1 of || j,k xj,k πj,k || (the expectation being taken with respect to
the normalized counting measure on the group of all signed permutation matrices
(πj,k ) i.e. πj,k ∈ {−1, 0, 1}) are equivalent. Conversely, if the L1 -norm and the
2
norm associated with ψ2 (t): = et − 1 are equivalent, then the volume of Dn must
satiesfy the above inequality up to some logarithmic factor. We prove that such
an inequality can not hold. We also include a proof of an upper estimate for the
volume of a convex polytope all of whose vertices are at a given distance from the
origin. Though this result is known we could not find a reference.
Received December 11, 1991; revised October 1, 1992.
1980 Mathematics Subject Classification (1991 Revision). Primary 15A51; Secondary 52A22.
190
M. SCHMUCKENSCHLÄGER
2
It is easy to see that the subspace E of Rn defined by
E=
2n n
o
\
2
x ∈ Rn : hx, Nj i = 0
j=1
where
N1 = (1, . . . , 1, 0, . . . , 0),
| {z }
n
N2 = (0, . . . , 0, 1, . . . , 1, 0, . . . , 0), . . .
| {z } | {z }
n
n
Nn = (0, . . . , 0, 1, . . . , 1),
| {z }
n
Nn+1 = (1, 0, . . . , 0, 1, 0, . . . , 0, 1, . . . ),
| {z } | {z }
n−1
n−1
Nn+2 = (0, 1, 0, . . . , 0, 1, 0, . . . , 0, 1, . . . ), . . .
| {z } | {z }
n−1
n−1
N2n = (0, . . . , 0, 1, 0, . . . , 0, 1, . . . )
| {z } | {z }
n−1
n−1
has dimension (n − 1)2 . Thus the dimension of Dn is (n − 1)2 .
2. The Basic Estimates
In order to estimate the (n−1)2 -dimensional volume of Dn we need some results.
The first one is due to Vaaler [V] (a generalization of this result can be found in
[M-P]).
Lemma 2.1. Let E be a k-dimensional subspace of Rn . Then
Volk (Bn∞ ∩ E) ≥ 2k
where Bn∞ is the cube [−1, 1]n.
The next result is a classical inequality of Urysohn (for an elementary proof we
refer to [P]).
Lemma 2.2. Let B be a convex symmetric body in Rn . Then
Voln (B)
Voln (Bn2 )
n1
Z
≤
S n−1
||x||2B ∗
12
dλ(x)
ON THE VOLUME OF THE DOUBLE STOCHASTIC MATRICES
191
where Bn2 is the unit ball of `2n , B ∗ is the polar of B and λ is the normalized
Lebesgue measure on S n−1 .
It is well known that the latter integral can be expressed as a gaussian integral,
i.e.
Z
12
Z
12
1
||x||2B ∗ dλ(x)
=√
||x||2B ∗ dγn (x)
n
S n−1
Rn
where γn is the canonical gaussian probability measure on Rn .
Lemma 2.3. Let g1 , . . . , gk be not necessarily independent gaussian variables
with mean zero. Then
! 12
2
E sup |gj |
j≤k
1
≤ c(1 + log k) 2 sup ||gj ||2
j≤k
For a proof we refer to [P].
Lemmas 2.2 and 2.3 immediately imply the following
Proposition 2.4. Let x1 , . . . , xk be unit vectors in Rn . Then
Voln (co(x1 , . . . , xk )
Voln (Bn2 )
n1
r
≤c
log k
n
Proof. By Lemma 2.2 we have
Voln (B)
Voln (Bn2 )
n1

X
 12
1
≤ √ E||
gj ej ||2B ∗ 
n
j=1
where B is the absolutely convex hull of {x1 , . . . , xn } and (gj )nj=1 are independent
standard gaussian variables. Since
||x||B ∗ = sup |hxi , xi|
i≤k
we get from Lemma 2.3
E||
X
j=1
gj ej ||2B ∗ = E sup |
i≤k
X
gj hxi , ej i|2
j=1
≤ c(1 + log k)
192
M. SCHMUCKENSCHLÄGER
Theorem 2.5. There exists an absolute constant c such that the following inequalities hold.
1
2
cp
≤ Vol(n−1)2 (Dn ) (n−1)2 ≤
log n
n
n
Proof. Let P0 be the n × n matrix with the constant entry n1 in each place.
Then we conclude by Lemma 2.1
2
Vol(n−1)2 (Dn ) = Vol(n−1)2 [0, 1]n ∩ (E + P0 )
!
n2
1
1
= Vol(n−1)2
− ,1 −
∩E
n
n
!
n2
1 1
≥ Vol(n−1)2
− ,
∩E
n n
(n−1)2
2
≥
n
Thus the left hand side inequality is established. As for the right hand side observe
that for any permutation matrix Π:
Π − P0 ||HS =
2 ! 12
√
1
1
(n − n) 2 + n 1 −
= n−1
n
n
2
Since the number of permutation matrices is n! we deduce from the above proposition
s
! 1 2
(n−1)
p
√
Vol(n−1)2 (Dn )
log n!
≤ n − 1c
≤ c1 log n
2
2
2
Vol(n−1) (B(n−1)2 )
(n − 1)
Hence
1
(V ol Dn ) (n−1)2 ≤
c2 p
log n
n
References
[M-P] Meyer M. and Pajor A., Section of the unit ball of `pn , Journal of functional analysis 80
(1988).
[P]
Pisier G., The volume of convex bodies and Banach space geometry, Cambridge, 1989.
[S]
Schmuckenschläger M., Kahane type inequalities for subgroups of the orthogonal group,
preprint.
Vaaler D., A geometric inequality with applications to linear forms, Pacific J. Math. 83
[V]
(1979).
M. Schmuckenschläger, Johannes Kepler Universität, 4040 Linz, Austria